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MathRevolution
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2! + \(\frac{5!}{3!2!}\)

= 12
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=>

If x is an odd number, then f(x) is an even number.
If x is an even number, then f(x) is an odd number.
Then the possible values of f(1) and f(3) are 2 and 4. The number of possible cases for f(1) and f(3) is 2.
The possible values of f(2) are 1, 3, and 5. The number of possible cases for f(2) = 3.

Thus. the number of possible cases for the function f(x) is 2·2·3 = 12.

Therefore, D is the correct answer.
Answer: D
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MathRevolution
=>

If x is an odd number, then f(x) is an even number.
If x is an even number, then f(x) is an odd number.
Then the possible values of f(1) and f(3) are 2 and 4. The number of possible cases for f(1) and f(3) is 2.
The possible values of f(2) are 1, 3, and 5. The number of possible cases for f(2) = 3.

Thus. the number of possible cases for the function f(x) is 2·2·3 = 12.

Therefore, D is the correct answer.
Answer: D

Why should we calculate the product here, MathRevolution?

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